An ideal voltmeter connected across a certain fresh 9-V battery reads 9.30 V, and an ideal ammeter briefly connected across the same battery reads 3.70 A. We say the battery has an open-circuit voltage of 9.30 V and a short-circuit current of 3.70 A. Model the battery as a source of emf ε in series with an internal resistance r as in Figure 27.1a. Determine both (a) ε and (b) r. An experimenter connects two of these identical batteries together as shown in Figure P27.45. Find (c) the open-circuit voltage and (d) the short-circuit current of the pair of connected batteries. (e) The experimenter connects a 12.0-Ω resistor between the exposed terminals of the connected batteries. Find the current in the resistor. (f) Find the power delivered to the resistor. (g) The experimenter connects a second identical resistor in parallel with the first. Find the power delivered to each resistor. (h) Because the same pair of batteries is connected across both resistors as was connected across the single resistor, why is the power in part (g) not the same as that in part (f)?
Figure P27.45
(a)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
In an open circuit current the current of the battery is
Formula to calculate the emf of the battery is,
Here,
Substitute
Conclusion:
Therefore, the emf of the battery is
(b)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
Formula to calculate the internal resistance of the battery is,
Here,
Substitute
Conclusion:
Therefore, resistance of the battery is
(c)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
Formula to calculate the total emf of the battery is,
Here,
Substitute
The total emf of the battery is equal to the open circuit voltage of the battery.
Conclusion:
Therefore, the open circuit voltage of the battery is
(d)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
The total resistance in the battery is,
Here,
Substitute
Thus, the internal resistance of the battery is
Formula to calculate the short circuit current of the batteries is,
Here,
Substitute
Conclusion:
Therefore, the short circuit current of the pair of connected batteries is
(e)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
The total series resistance in the battery is,
Here,
Substitute
Thus, the total series resistance of the battery is
Formula to calculate the current in the resistor
Here,
Substitute
Conclusion:
Therefore, the current in the resistor
(f)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
Formula to calculate the power delivered to the resistor is,
Here,
Substitute
Conclusion:
Therefore, the power delivered to the resistor is
(g)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
The batteries are connected in series. The voltages of the both batteries are same.
The equivalent internal resistance in the battery is,
Here,
Substitute
Thus, the total resistance of the resistor is
Formula to calculate the current in the batteries is,
Here,
Substitute
Thus, the current produced in the batteries is
Formula to calculate the terminal voltage across both batteries is,
Here,
Substitute
Thus, the terminal voltage across both batteries is
Formula to calculate the power delivered to each resistor is,
Here,
Substitute
Conclusion:
Therefore, the power delivered to each resistor is
(h)
Answer to Problem 28.74AP
Explanation of Solution
Given info: The open circuit voltage of the battery is
In part (g), the total internal resistance of the resistor is
Conclusion:
Therefore, the internal resistance of the batteries and the terminal voltage of the batteries is not same in both cases.
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Chapter 28 Solutions
Physics for Scientists and Engineers
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